10. (a) Use the substitution $x = u^2 + 1$ to show that \[ \int_1^0 \frac{3 \, dx}{(x-1)(3 + 2\sqrt{x - 1})} = \int_0^q \frac{6 \, du}{u(3 + 2u)} \] where $p$ and $q$ are positive constants to be found - Edexcel - A-Level Maths Pure - Question 12 - 2020 - Paper 1

Using the result from part (a), we continue from:

$\int_0^{-1} \frac{6 \, du}{u(3 + 2u)}.$

We can break this down using partial fractions:

$\frac{6}{u(3 + 2u)} = \frac{A}{u} + \frac{B}{3 + 2u}.$

Solving for $A$ and $B$, we find:

$6 = A(3 + 2u) + Bu \Rightarrow 6 = 3A + (2A + B)u.$

Setting coefficients equal gives:

1. $3A = 6$
2. $2A + B = 0$

From the first equation, we obtain $A = 2$. Substituting into the second:

$2(2) + B = 0 \Rightarrow B = -4.$

Thus, we can rewrite the integral:

$\int \left( \frac{2}{u} - \frac{4}{3 + 2u} \right) du = 2 \ln |u| - 2\ln |3 + 2u| + C.$

Evaluating from $0$ to $-1$, we find:

$= \ln a,$

where $a = \frac{16}{9}$. Thus, we conclude with the final integral equating to:

$\ln \frac{16}{9}.$